STRUCTURE OF SILICON ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ”(2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Silicon (Si) has 24 known isotopes, with mass numbers ranging from 22 to 45. Si-28 (the most abundant isotope, at 92.23%), Si-29 (4.67%), and Si-30 (3.1%) are stable. The longest-lived radioisotope is Si-32, which is produced by cosmic ray spallation of argon. Its half-life has been determined to be approximately 170 years (0.21 MeV), and it decays by beta emission to P-32 (which has a 14.28 day half-life 1) and then to S-32. After Si-32, Si-31 has the second longest half-life at 157.3 minutes. All others have half-lives under 7 seconds. The standard atomic mass is 28.0855(3) u. The least stable is usually Si-43 with a half-life greater than 60 nanoseconds. ' ' WHY Si-28, Si-29 AND Si-30 ARE STABLE NUCLIDES ' After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. For understanding the stability of Si-28 with S = 0 you can read my STRUCTURE OF Mg-27, Al-27 AND Si-28 . Using the following diagram of Si-29 we see that the structure of Si-28 is a stable nuclide because outside the core (Mg-24 with S=0 )the nucleons like the p13, n13, p14 and n14 make vertical rectangles with strong vertical bonds with S=0. Thus the stability of Si-28 is due to these two rectangles with S =0. Also the Si-29 with S =+1/2 is a stable nuclide, because the extra neutron , the n, of positive spin makes one horizontal bond with p10 and a second strong vertical bond with p14. ' DIAGRAM OF THE STABLE Si-29 WITH S = +1/2 ' ' ' p12......n12' ' -HP6 ''' '' n11......p11 '' '' ' n10.......p10 n' ' +HP5 p9.........n9 ' ' p8........n8.......p14' ' -HP4 p13.....n7........p7 ' ' n6........p6.......n14' ' +HP3 n13......p5........n5 ' ' p4........n4' ' -HP2 n3........p3 ' ' n2........p2' ' +HP1 p1........n1 ' ' ' ' '''However the structure of Si-30 is very different. In the following diagram of Si-30 with S=0 we see that the p13, n13, p14 and n14 make a vertical rectangle of S=0 which is stable because the two extra neutrons o f opposite spins, the 2n make two bonds per neutron able to overcome the nn repulsions of short range. They also contribute to the increase of the binding energies of the horizontal bonds able to overcome the pp repulsions of long range. Especially the n of positive spin makes a horizontal bond with p10 at +HP5 and a strong vertical bond with p14. Moreover the n of negative spin makes a horizontal bond with p3 at -HP2 and a strong vertical bond with p13. ' ' '' ''' DIAGRAM OF THE STABLE Si-30 WITH S=0''' ' ' ' p12......n12' ' -HP6 n11......p11 ' ' n10.......p10 n' ' +HP5 p9.........n9 ' ' p8........n8.......p14' ' -HP4 n7........p7.........n13 ' ' n6........p6.......n14' ' +HP3 p5........n5.......p13 ' ' p4........n4' ' -HP2 n3........p3 n ' ' n2........p2' ' +HP1 p1........n1 ' ' ' NUCLEAR STRUCTURE OF Si-32, Si-34, Si-36, Si-38, Si-40, Si-42 AND Si-44 WITH S =0 After a careful analysis of the above nuclides we discovered that the structure of them is based on the structure of Si-30, because the extra neutrons of opposite spins make weak horizontal bonds with the protons of the core. For example using s the diagram of S-30 we see that the S-44 with S= 0 has 14 extra neutrons of opposite spins (more than those of S-30 ) which give a total S=0 because they make weak horizontal bonds with the protons of the core. NUCLEAR STRUCTURE OF Si-31 AND Si-33 WITH S =+3/2 For understanding better the structure of Si-31 you can read my STRUCTURE OF Si-31 AND P-31 . In the following diagram of Si-31 you see that the two extra neutrons, the 2n, of positive spins make two bonds per neutron and exist at the +HP5 giving a total spin S = +1. In this case the third extra neutron the (n) of positive spin S =+1/2 exists at the +Hp1. Thus we get S = +1 +1/2 = +3/2 Similarly the Si-33 has the same S =+3/2 because there are two more extra neutrons of opposite spins making single horizontal bonds with the protons of the core. ' ' ' ' ' DIAGRAM OF Si-31 WITH S =+3/2 ' ' ' ' p12......n12' ' -HP6 ''' ''n11......p11 '' '' ' n10.......p10 n' ' +HP5 n p9.........n9 ' ' p8........n8.......p14' ' -HP4 p13.....n7........p7 ' ' n6........p6.......n14' ' +HP3 n13......p5........n5 ' ' p4........n4' ' -HP2 n3........p3 ' ' n2........p2' ' +HP1 (n) p1........n1 ' ' ' NUCLEAR STRUCTURE OF Si-27, Si-25 AND Si-23 In the following diagram of Si-27 you see that the core is a parallelepiped with 5 horizontal planes like the +HP1, the -HP2, the +HP3 , the -HP4 and the +Hp5 giving S = +4. Here the nucleons like the p11, n11, p12, n12, p13, n13 and p14 make vertical rectangles with S= +1/2. Thus we have 14 protons and 13 neutrons with a total spin S =+5/2. Then in the absence of two neutrons of opposite spins like the n1 of the +Hp1 and the n12 of the -HP2 we get the same S =+5/2. In the case of Si-23 with S =+3/2 the two more absent neutrons of positive spins give S=+3/2 because S = +5/2 - 2(+1/2) = +3/2. ' ' ' DIAGRAM OF Si-27 WITH S = +5/2' ' p10..........n10.....p14' ' +HP5 p12.....n9...........p9 ' ' n8.........p8.......n13 ' ' -HP4 n11.....p7..........n7 ' ' p6........n6.......p13' ' +HP3 p11.....n5..........p5 ' ' n4.........p4........n12' ' -HP2 p3..........n3 ' ' p2........n2' ' +HP1 n1........p1 ' NUCLEAR STRUCTURE OF Si-26, Si-24 AND Si-22 WITH S =0 After a careful analysis of the above nuclides we discovered that their structure is based on the structure of Si-28 with S=0. For example in the absence of two neutrons of opposite spins like the n1 and n12 we get the same spin S = 0 as that of Si-28. ' ' 'NUCLEAR STRUCTURE OF Si-35, Si-37, Si-39 AND Si-41 WITH S = -7/2 ' After a careful analysis of these nuclides I discovered that their structure is based on the structure of Si-27 since the nucleons change the spins. For example using the diagram of Si-27 we see that the -HP1, the +HP2, the -HP3 , the +HP4 and the -HP5 give a total spin S=-5/2. Then adding two extra neutrons of negative spins and 6 extra neutrons of opposite spins we get the structure of Si-35 with S =-7/2. In the same way adding two neutrons of opposite spins in te structure of Si-35 we get the structure of Si-37 with S =-7/2. Category:Fundamental physics concepts